Have a query? Ask a Tutor!!

Access to best educators and exclusive paper discussions

Offsite Campus Turito Championship

Can’t find what you’re looking for?

Our tutors are from top schools around the world, and they are waiting for you 24/7, anytime, anywhere.

Homework- One to One Solutions
Understand concepts to solve better
Get your doubts solved instantly

Physics-

General

Easy

Question

A TV tower has a height of 100 m. How much population is covered by the TV broadcast if the average population density around the tower is $100 blank k m to the power of negative 2 end exponent$ (radius of the earth $equals 6.37 cross times 10 to the power of 6 end exponent$ m)

4 lakh
4 billion
40,000
40 lakh

The correct answer is: 40 lakh

$d equals square root of 2 h R end root$
Population covered
$equals pi d to the power of 2 end exponent cross times$ population density
$equals 3.114 cross times left parenthesis 2 cross times 0.1 cross times 6.37 cross times 10 to the power of 3 end exponent right parenthesis cross times 1000 almost equal to 40$ lakh

Related Questions to study

The temperature variation in the region of stratosphere lies from

The temperature variation in the region of stratosphere lies from

Physics-General

The amplitude of the magnetic field part of a harmonic Electromagnetic Wave in vacuum is B₀=510 nT. What is the amplitude of the electric field part of the wave?

The amplitude of the magnetic field part of a harmonic Electromagnetic Wave in vacuum is B₀=510 nT. What is the amplitude of the electric field part of the wave?

Physics-General

The value of $3 to the power of log subscript 4 end subscript invisible function application 5 end exponent minus 5 to the power of log subscript 4 end subscript invisible function application 3 end exponent$ is

The value of $3 to the power of log subscript 4 end subscript invisible function application 5 end exponent minus 5 to the power of log subscript 4 end subscript invisible function application 3 end exponent$ is

parallel

The number of distinct real roots of $open vertical bar table row cell sin invisible function application x end cell cell cos invisible function application x end cell cell cos invisible function application x end cell row cell cos invisible function application x end cell cell sin invisible function application x end cell cell cos invisible function application x end cell row cell cos invisible function application x end cell cell cos invisible function application x end cell cell sin invisible function application x end cell end table close vertical bar equals 0$ in the interval $– fraction numerator pi over denominator 4 end fraction less or equal than x less or equal than fraction numerator pi over denominator 4 end fraction$ is

The number of distinct real roots of $open vertical bar table row cell sin invisible function application x end cell cell cos invisible function application x end cell cell cos invisible function application x end cell row cell cos invisible function application x end cell cell sin invisible function application x end cell cell cos invisible function application x end cell row cell cos invisible function application x end cell cell cos invisible function application x end cell cell sin invisible function application x end cell end table close vertical bar equals 0$ in the interval $– fraction numerator pi over denominator 4 end fraction less or equal than x less or equal than fraction numerator pi over denominator 4 end fraction$ is

A piece of paper is in the shape of a square of side 1 m long. It is cut at the four corners to make a regular polygon of eight sides (octagon). The area of the polygon is

A piece of paper is in the shape of a square of side 1 m long. It is cut at the four corners to make a regular polygon of eight sides (octagon). The area of the polygon is

The number of solution of the equation $tan invisible function application x tan invisible function application 4 x equals 1 blank$ for $0 less than x less than pi$ is

The number of solution of the equation $tan invisible function application x tan invisible function application 4 x equals 1 blank$ for $0 less than x less than pi$ is

parallel

The value of $cot invisible function application 70 degree plus 4 cos invisible function application 70 degree$ is

The value of $cot invisible function application 70 degree plus 4 cos invisible function application 70 degree$ is

In triangle $A B C comma blank angle A equals 60 degree comma angle B equals 40 degree blank$ and $blank angle C equals 80 degree.$ If $P$ is the centre of the circumcircle of triangle $A B C$ with radius unity, then the radius of the circumcircle of triangle $B P C$ is

In triangle $A B C comma blank angle A equals 60 degree comma angle B equals 40 degree blank$ and $blank angle C equals 80 degree.$ If $P$ is the centre of the circumcircle of triangle $A B C$ with radius unity, then the radius of the circumcircle of triangle $B P C$ is

If in $blank increment A B C comma sin invisible function application A cos invisible function application B equals fraction numerator square root of 2 minus 1 over denominator square root of 2 end fraction$ and $sin invisible function application B cos invisible function application A equals fraction numerator 1 over denominator square root of 2 end fraction comma blank$ then the triangle is

If in $blank increment A B C comma sin invisible function application A cos invisible function application B equals fraction numerator square root of 2 minus 1 over denominator square root of 2 end fraction$ and $sin invisible function application B cos invisible function application A equals fraction numerator 1 over denominator square root of 2 end fraction comma blank$ then the triangle is

parallel

The equation $2 cos to the power of 2 end exponent invisible function application fraction numerator x over denominator 2 end fraction sin to the power of 2 end exponent invisible function application x equals x to the power of 2 end exponent equals x to the power of negative 2 end exponent semicolon 0 less than x less or equal than fraction numerator pi over denominator 2 end fraction$ has

The equation $2 cos to the power of 2 end exponent invisible function application fraction numerator x over denominator 2 end fraction sin to the power of 2 end exponent invisible function application x equals x to the power of 2 end exponent equals x to the power of negative 2 end exponent semicolon 0 less than x less or equal than fraction numerator pi over denominator 2 end fraction$ has

If $sin invisible function application x plus c o s e c blank x equals 2 comma$ then $sin to the power of n end exponent invisible function application x plus blank c o s e c to the power of n end exponent x$ is equal to

If $sin invisible function application x plus c o s e c blank x equals 2 comma$ then $sin to the power of n end exponent invisible function application x plus blank c o s e c to the power of n end exponent x$ is equal to

In triangle $A B C comma blank$ if $tan invisible function application open parentheses A divided by 2 close parentheses equals 5 divided by 6$ and $tan invisible function application open parentheses B divided by 2 close parentheses equals 20 divided by 37 comma blank$ the sides $a comma blank b blank$ and $blank c$ are in

In triangle $A B C comma blank$ if $tan invisible function application open parentheses A divided by 2 close parentheses equals 5 divided by 6$ and $tan invisible function application open parentheses B divided by 2 close parentheses equals 20 divided by 37 comma blank$ the sides $a comma blank b blank$ and $blank c$ are in

parallel

$cos to the power of 3 end exponent invisible function application x sin invisible function application 2 x equals not stretchy sum from x equals 0 to n of a subscript r end subscript sin invisible function application open parentheses r blank x close parentheses for all x element of R comma blank t h e n$

$cos to the power of 3 end exponent invisible function application x sin invisible function application 2 x equals not stretchy sum from x equals 0 to n of a subscript r end subscript sin invisible function application open parentheses r blank x close parentheses for all x element of R comma blank t h e n$

If the equation $cot to the power of 4 end exponent invisible function application x minus 2 blank c o s e c to the power of 2 end exponent blank x plus a to the power of 2 end exponent equals 0$ has at least one solution, then the sum of all possible integral values of $´ a to the power of ´ end exponent$ is equal to

If the equation $cot to the power of 4 end exponent invisible function application x minus 2 blank c o s e c to the power of 2 end exponent blank x plus a to the power of 2 end exponent equals 0$ has at least one solution, then the sum of all possible integral values of $´ a to the power of ´ end exponent$ is equal to

Let $blank f open parentheses n close parentheses equals 2 cos invisible function application n x for all n element of N comma blank t h e n blank f open parentheses 1 close parentheses blank f open parentheses n plus 1 close parentheses minus f open parentheses n close parentheses$ is equal to

Let $blank f open parentheses n close parentheses equals 2 cos invisible function application n x for all n element of N comma blank t h e n blank f open parentheses 1 close parentheses blank f open parentheses n plus 1 close parentheses minus f open parentheses n close parentheses$ is equal to

parallel

card img

With Turito Academy.

card img

With Turito Foundation.

card img

Get an Expert Advice From Turito.

Turito Academy

card img

With Turito Academy.

Test Prep

card img

With Turito Foundation.